Let LL^{prime} be the latus rectum through th | vedclass

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(A) Let the equation of the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. The coordinates of the vertex $A^{\prime}$ are $(-a, 0)$. The focus

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$\frac{\sqrt{3}+1}{\sqrt{3}}$

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(A) Let the equation of the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. The coordinates of the vertex $A^{\prime}$ are $(-a, 0)$. The focus $S$ is $(ae, 0)$ and the endpoints of the latus rectum $L$ and $L^{\prime}$ are $(ae, \frac{b^2}{a})$ and $(ae, -\frac{b^2}{a})$. The length of the side $LL^{\prime} = \frac{2b^2}{a}$. Since $	riangle A^{\prime}LL^{\prime}$ is equilateral,the height from $A^{\prime}$ to $LL^{\prime}$ is $\frac{\sqrt{3}}{2} 	imes 	ext{side length} = \frac{\sqrt{3}}{2} 	imes \frac{2b^2}{a} = \frac{\sqrt{3}b^2}{a}$. The distance from $A^{\prime}(-a, 0)$ to the line $x = ae$ is $ae - (-a) = a(e+1)$. Thus,$a(e+1) = \frac{\sqrt{3}b^2}{a}$. Using $b^2 = a^2(e^2-1)$,we get $a(e+1) = \frac{\sqrt{3}a^2(e^2-1)}{a} = \sqrt{3}a(e-1)(e+1)$. Dividing by $a(e+1)$,we get $1 = \sqrt{3}(e-1)$,which implies $e-1 = \frac{1}{\sqrt{3}}$,so $e = 1 + \frac{1}{\sqrt{3}} = \frac{\sqrt{3}+1}{\sqrt{3}}$.

Let $LL^{\prime}$ be the latus rectum through the focus $S$ of a hyperbola and $A^{\prime}$ be the opposite vertex of the hyperbola. If triangle $A^{\prime}LL^{\prime}$ is equilateral,then the eccentricity of the hyperbola is

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